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9:41 AM
Man there's a theory of gravity for just about every Cartan geometry above $SO(3,1)$
 
9:57 AM
What's a good pair of letters for two curves
I don't want to use $\gamma, \gamma'$ because it looks a little like a derivative
Using numbers may look like coordinate indices!
 
I don't think $\gamma_1,\gamma_2$ are so easily mistaken for coordinate indices
 
10:13 AM
Ah well let's go with it
Maybe we should start using more alphabets
I bet some hieroglyphs would be good for that
𓆓
He is a little curve
looks like Chrome doesn't like hieroglyphs
 
just start using Japanese Hiragana and Katakana :P
together that's almost 100 additional symbols I think
 
yeah but they're not very representative
 
10:41 AM
One thing that is unfortunate is that it's pretty hard to tell the contact of two curves at a glance beyond the first one
I can't say if two curves share the same second derivative like that
 
glS
11:04 AM
how about ქართული ენა
always thought Georgian has a pretty awesome script
 
@Slereah dunno, I'd love calling a variable シ
 
"Perspectives in Geometry and Relativity, Essays in honour of Vaclav Hlavaty"
God I hate those books
They're up there with "Conference books" on the list of bad physics books
That one's on libgen at least
If I die don't write any essays for me
 
11:25 AM
not like you could do anything about it at that point :P
 
I will haunt you
With my ghost and antighost
This paper uses $\mathscr{G} \Xi(M)$
The fanciest of notation
 
hello everyone, what is the dimension(number of generators) of the product of two symmetry groups? for example, I have $U(N)$ of dimension $N^2$ and $SU(N)$ of dimension $N^2-1$. What is the dimension of $U(N)\times SU(N)$? is the sum of the dimensions of the two groups?
 
If it's a direct product I think so yes
Same as the product of two manifolds
The algebra of a product of two Lie groups is their direct sum
so $\mathfrak{u}(N) \oplus \mathfrak{su}(N)$
 
@Slereah yes it's a direct product
thank you!
 
11:56 AM
anyone ever dealt with a lagrangian with $U(N)\times SU(N)$ symmetry? do you have any references?
 
Don't think I've seen that specific group, no
 
12:37 PM
of course n-flavour QCD has $SU(N)\times SU(N)\times U(1)$, which is almost $SU(N)\times U(N)\cong SU(N)\times (SU(N)\rtimes U(1))$
the semidirect product can be converted into a direct one only if you are willing to mod out by a $\mathbb Z_N$
 
1:35 PM
folks
I need a quick external check
0
Q: Why $S_z|+⟩=\tfrac{\hbar}{2} |+⟩$

KashmiriI'm studying Sakurai and I've got a basic question. Here is the relevant paragraph: One of the physics postulates is that $|\alpha\rangle$ and $c|\alpha\rangle$, with $c \neq 0$, represent the same physical state. An observable, such as momentum and spin components, can be represented by an oper...

Is the title of this question displaying erroneously on the front page for you?
I see this
but the MathJax in the title looks fine to me
 
WOW, that's an odd one. I'll see if I can't reproduce this in my local environment and get a fix out for it. Thanks for the report! — Ben Kelly ♦ yesterday
 
@Wolgwang uuufffffffff
wow
 
glS
1:54 PM
even mathjax can't stand the new design lol
 
@EmilioPisanty looks good to me
btw, there are any plugins to show latex text in this chat graphically and not the actual code?
 
Jim
chatjax
 
34
A: Any chance of MathJax in chat?

Ilmari KaronenAs a workaround while this request is pending, there exist several client-side workarounds that can be used to enable LaTeX rendering in chat, including: ChatJax, a set of bookmarklets by robjohn to enable dynamic MathJax support in chat. Commonly used in the Mathematics chat room. An alternat...

 
i've installed greasemonkey and chatjax++ but still see the latex code, even after refresh
 
glS
2:12 PM
so, upon some further reflections on the stuff about the tautological 1-form of yesterday. Effectively, the tautological 1-form can be thought of as attaching a number to every triple of (1) basepoint $q\in Q$, (2) covector $m\equiv(q,p)\in T_q^* Q$ attached to $q$, and (3) tangent vector $v\equiv(q,\tilde v)\in T_q Q$ attached to $q$. So why define this mapping as some $\theta:T^*Q\to T^*(T^* Q)$, or as $\theta:T(T^* Q)\to\mathbb R$?

What I mean is, why bother double (co)tangent bundles for this? Isn't it easier to understand it if we define it as a "metric-like object",mapping each $q\in
when we define it via $\theta_m=m\circ d\pi_m$ (in the wikipedia page notation), we have to deal with $TT^* Q$ due to $d\pi$.. but then we immediately discard the nontrivial "vertical" piece of said tangent bundle. So why are we mentioning it at all, instead of just defining the object using the simpler $TQ$ and $T^* Q$?
 
2:30 PM
@glS Can you write down what sort of mathematical object your "metric-like object" is without mentioning a double (co)tangent bundle?
 
@john Someone here might be able to help you...
 
@Wolgwang ok thanks
 
glS
@ACuriousMind isn't $\theta_q(p,\tilde v)=p(\tilde v)$ a good enough definition? Or you mean to write it as some $\theta_q=(...)$? In such case... can't we use notation like those used for metrics? We write $g=g_{ij} dx_i dx_j\equiv g_{ij} dx_i dx_j$ to denote maps $g_p(v,w)=g_{ij}(p) \, v_i w_j$. Why not do the same here? Though admittedly, we'd have to use the double dual in this case? Would that be weird?
so we'd have something like $\theta=dq^i dp_i$, with $\theta$ being then, I think, a section of the bundle $T^*Q\otimes T^{**}Q\to Q$. Because we need to talk about the functionals defined on functionals
never seen the dual of the dual used in this context though, so maybe there's a reason for that
it just bothers me that the standard definition uses $T^* T^* Q$, but then actually we just use $m\in T_q^* Q$ and tangent vectors on its basepoint. Never seemingly actually making use of the complexity of $T^* T^* Q$
 
2:55 PM
a pretty cool paper
""Ha!" cries the mathematician, as keen in pursuit of a problem as a greyhound in pursuit of a hare, "I spy partial differential equations! Let's solve them!" And before you can say "Jack Robinson", away he goes in hot pursuit, $T_{ab}$ given (Of course! The right-hand side of an equation is always given!) and $g_{ab}$ to be found."
"Why should we assume one known, rather than either of the others ? The greyhound returns to leash, a sadder and a wiser dog."
 
@glS I think I understand the issue - you're essentially asking why we want to think of $\theta$ as being a 1-form on $T^\ast Q$ (and hence a section of $T^\ast T^\ast Q$ or whatever)
A very simple answer is that we want to be able to write $\omega = \mathrm{d}\theta$ for the symplectic form on $T^\ast Q$
and since $\omega$ is a 2-form on the cotangent bundle, $\theta$ is the 1-form
really, the main purpose of the tautological one form is to allow us to define the symplectic structure on the cotangent bundle - the "replace velocities with momenta" is a nice interpretation, but nobody ordered that :P
I'm not convinced there's any more meaning to be squeezed out of it than that
 
glS
3:16 PM
@ACuriousMind mh. So, we want to define a notion of "area" in the cotangent bundle/phase space, which we'd do in the naive way via $dq^i\wedge dp_i$, and then we observe that this can be written as the exterior derivative of some one-form, which is the tautological one-form
and because we want "areas" on $T^* Q$, we want 2-forms $TT^*Q\wedge TT^*Q \to\mathbb R$, which can be seen as induced by 1-forms $TT^*Q\to\mathbb R$. Ok, this way of looking at it makes sense
 
3:43 PM
You know I wonder if a nice way to settle the whole "diffeomorphism generates GR" is just to look at other G-structure theories
I'm pretty sure that all the whatever gravities are about as much generated by the diffeomorphism group as GR
Nobody's gonna claim that the Thomas-Whitehead gravity is the fundamental theory generated by diffeomorphisms
 
4:39 PM
Is there a gravity theory if the structure group is just $e$ I wonder
Only a single frame is the True Frame
Although I guess that as a theory, it's not very dynamic
Not gonna get a very large connection from a zero dimensional bundle
Connection form is just gonna be $\omega: M \to e$
Action is just zero
And a cosmological constant I guess
 
I have a Hamiltonian $H$ with electron and phonon states. The notes then say we integrate out the phonon modes to consider an effective electron Hamiltonian? What does this mean? Why can we integrate out something?
 
Generated by the determinant of the One True Frame
$$S = \int e$$
Well e is the frame here, not the neutral element
 
 
1 hour later…
6:12 PM
I'm reading a paper on Gas Hydrates and I'm confused by a line. Hoping someone can clarify?
Loveday 2007: "The importance of repulsions to clathrate stability can be seen from the fact that neither water itself nor ammonia, which readily forms hydrogen bonds to water, adopt these cage structures in spite of the fact that both ammonia and water have sizes comparable with those of clathrate-forming guests"
 
 
2 hours later…
8:02 PM
Seems more like a chemistry chat question
 

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